Then, since BE is equal to ED, for E is the centre, and EA is common and at right angles, therefore the base AB is equal to the base AD. [[I. 4](elem.1.4)]

For the same reason each of the straight lines BC, CD is also equal to each of the straight lines AB, AD; therefore the quadrilateral ABCD is equilateral.

I say next that it is also right-angled.

For, since the straight line BD is a diameter of the circle ABCD, therefore BAD is a semicircle; therefore the angle BAD is right. [[III. 31](elem.3.31)]

For the same reason each of the angles ABC, BCD, CDA is also right; therefore the quadrilateral ABCD is right-angled.

But it was also proved equilateral; therefore it is a square; [[I. Def. 22](elem.1.def.22)] and it has been inscribed in the circle ABCD.

Therefore in the given circle the square ABCD has been inscribed. Q. E. F.